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Conservative forces Suppose a force F acting on a particle depends only on the position r of the particle, i.e. F = F(r). The work W done by F in moving the particle from r 1 to r 2 is defined to be If W does not depend on the choice of a path for r, then the force F is called conservative. Where does this name come from? r1r1 r2r2

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Conservative forces If F is conservative, then where the color framing the integral indicates the path taken from r 1 to r 2. r1r1 r2r2

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r1r1 r2r2 Conservative forces If F is conservative, then where the color framing the integral indicates the path taken from r 1 to r 2.

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Conservative forces If F is conservative, then we can define a potential energy U(r) such that U(r 2 ) – U(r 1 ) = and U(r) is a single-valued function of r (i.e. U(r) is a scalar field). With U(r) we can define a conserved total energy, the sum of the kinetic energy and the potential energy. r1r1 r2r2

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Conservation of energy If F is conservative, then we can define a potential energy U(r) such that U(r 2 ) – U(r 1 ) = and U(r) is a single-valued function of r (i.e. U(r) is a scalar field). With U(r) we can define a conserved total energy, the sum of the kinetic energy and the potential energy. Conclusion: a conservative force implies a conserved total energy. r1r1 r2r2

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Conservation of energy Is the electric force a conservative force? If not, then we can use an electric field to get energy for free. Here’s how:

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Conservation of energy Is the electric force a conservative force? If not, then we can use an electric field to get energy for free. Here’s how: If the electric force F q on a particle of charge q is not conservative, then for some loop in r we can write FqFq FqFq FqFq

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Conservation of energy Is the electric force a conservative force? If not, then we can use an electric field to get energy for free. Here’s how: If the electric force F q on a particle of charge q is not conservative, then for some loop in r we can write which means that every time the charge makes a complete loop in r, it gets more kinetic energy! FqFq FqFq FqFq

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Electric potential energy We will prove that the electrostatic force is conservative! First, we will calculate the work W done by the electric force if we move a charge q 1 in a straight line towards a charge q 2. Then we will prove that W does not depend on the path.

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Electric potential energy We will prove that the electrostatic force is conservative! First, we will calculate the work W done by the electric force if we move a charge q 1 in a straight line towards a charge q 2. Then we will prove that W does not depend on the path. The charge q 2 is always at r 2 while the charge q 1 moves in a straight line to r 1 from a point infinitely far away. r2r2 r1r1

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Electric potential energy We will prove that the electrostatic force is conservative! First, we will calculate the work W done by the electric force if we move a charge q 1 in a straight line towards a charge q 2. Then we will prove that W does not depend on the path. The charge q 2 is always at r 2 while the charge q 1 moves in a straight line to r 1 from a point infinitely far away. r2r2 r1r1 r F(r) drdr

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Electric potential energy But now suppose the charge q 1 reached the point r 1 by the path indicated. What is the work done by the electric force? r2r2 r1r1 r

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r2r2 r1r1 F(r)F(r) r Electric potential energy But now suppose the charge q 1 reached the point r 1 by the path indicated. What is the work done by the electric force? drdr

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r2r2 r1r1 Electric potential energy But now suppose the charge q 1 reached the point r 1 by the path indicated. What is the work done by the electric force? r F(r)F(r) drdr

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r2r2 r1r1 Electric potential energy But now suppose the charge q 1 reached the point r 1 by the path indicated. What is the work done by the electric force? Since F(r) · dr is hard to compute … r F(r)F(r) drdr

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r2r2 r1r1 Electric potential energy But now suppose the charge q 1 reached the point r 1 by the path indicated. What is the work done by the electric force? Let ’ s resolve dr into dr ║ and dr ┴ : r F(r)F(r) drdr

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r2r2 r1r1 Electric potential energy But now suppose the charge q 1 reached the point r 1 by the path indicated. What is the work done by the electric force? Let ’ s resolve dr into dr ║ and dr ┴ : r F(r)F(r) drdr

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r2r2 r1r1 r2r2 r1r1 r F(r)F(r) Electric potential energy But now suppose the charge q 1 reached the point r 1 by the path indicated. What is the work done by the electric force? Resolved into dr ║ and dr ┴, the path looks like this …

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r2r2 r1r1 r2r2 r1r1 r F(r)F(r) Electric potential energy But now suppose the charge q 1 reached the point r 1 by the path indicated. What is the work done by the electric force? … which we arrange to look like this:

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r2r2 r1r1 r2r2 r1r1 r F(r)F(r) Electric potential energy But now suppose the charge q 1 reached the point r 1 by the path indicated. What is the work done by the electric force? And now we are back to the same integral we had before, since integration dr ┴ does not contribute anything:

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r2r2 r1r1 r2r2 r1r1 r F(r)F(r) Electric potential energy But now suppose the charge q 1 reached the point r 1 by the path indicated. What is the work done by the electric force? And now we are back to the same integral we had before, since integration dr ┴ does not contribute anything:

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Electric potential energy Conclusion 1: The electrostatic force is conservative. Conclusion 2: The electric potential energy of two charges q 1 and q 2 at points r 1 and r 2, respectively, is regardless of how the charges were assembled.

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Remember Faraday? We had some basic rules and observations about field lines: They never start or stop in empty space – they stop or start on a charge or extend to infinity. They never cross – if they did, a small charge placed at the crossing would show the true direction of the field there. The density of field lines in one direction is proportional to the strength of the field in the perpendicular direction.

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Remember Faraday? Now we can add a new basic rule about field lines: They never start or stop in empty space – they stop or start on a charge or extend to infinity. They never cross – if they did, a small charge placed at the crossing would show the true direction of the field there. The density of field lines in one direction is proportional to the strength of the field in the perpendicular direction. There are no closed loops! E E E E

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Remember Faraday? Now we can add a new basic rule about field lines: They never start or stop in empty space – they stop or start on a charge or extend to infinity. They never cross – if they did, a small charge placed at the crossing would show the true direction of the field there. The density of field lines in one direction is proportional to the strength of the field in the perpendicular direction. There are no closed loops! E E E E

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Electric potential energy What is the electric potential U(θ) energy of two charges q and –q separated by a rod of length d, when it tilts at an angle θ in a constant electric field E? (Assume U(90°) = 0.) FqFq θ F -q

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Electric potential energy What is the electric potential U(θ) energy of two charges q and –q separated by a rod of length d, when it tilts at an angle θ in a constant electric field E? (Assume U(90°) = 0.) Answer: FqFq θ F -q drdr

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Electric potential energy What is the electric potential U(θ) energy of two charges q and –q separated by a rod of length d, when it tilts at an angle θ in a constant electric field E? (Assume U(90°) = 0.) Answer: θ F -q drdr Not the same d! FqFq

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Electric potential energy What is the electric potential U(θ) energy of two charges q and –q separated by a rod of length d, when it tilts at an angle θ in a constant electric field E? (Assume U(90°) = 0.) Answer: θ F -q drdr FqFq

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Electric potential energy What is the electric potential U(θ) energy of two charges q and –q separated by a rod of length d, when it tilts at an angle θ in a constant electric field E? (Assume U(90°) = 0.) Answer: FqFq θ F -q ΔrΔr ΔrΔr

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Electric potential We have seen that “electric potential energy” can include all the charge in a system – for example, we calculated the total potential energy of a system of three point charges. Or it can mean the potential energy of only some of the charges in an external electric field – for example, we calculated the potential energy of a dipole in a constant electric field.

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Electric potential We have seen that “electric potential energy” can include all the charge in a system – for example, we calculated the total potential energy of a system of three point charges. Or it can mean the potential energy of only some of the charges in an external electric field – for example, we calculated the potential energy of a dipole in a constant electric field. In the case of one point charge in an external electric field, what we most often calculate is not the electric potential energy but rather the electric potential.

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Electric potential We have seen that “electric potential energy” can include all the charge in a system – for example, we calculated the total potential energy of a system of three point charges. Or it can mean the potential energy of only some of the charges in an external electric field – for example, we calculated the potential energy of a dipole in a constant electric field. In the case of one point charge in an external electric field, what we most often calculate is not the electric potential energy but rather the electric potential. The relation between the electric potential energy of a point charge vs. the electric potential at the position of the point charge is exactly analogous to the relation between the electric force and the electric field.

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Electric potential We have seen that “electric potential energy” can include all the charge in a system – for example, we calculated the total potential energy of a system of three point charges. Or it can mean the potential energy of only one point charge in an external electric field – for example, we calculated the potential energy of each charge in a dipole in a constant electric field. In the latter case – the case of a point charge in an external electric field – what we most often calculate is not the electric The relation between the electric potential energy of a point charge vs. the electric potential at the position of the point charge is exactly analogous to the relation between the electric force and the electric field. F q (r) is the electric force on a point charge q located at r. E(r) is the electric field at the point r. U(r) is the potential energy of a point charge q at r. V(r) is the electric potential at the point r.

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Electric potential The units of electric potential must be energy/charge. Our unit is the volt; 1 V = J/C = joules/coulombs. F q (r) is the electric force on a point charge q located at r. E(r) is the electric field at the point r. U(r) is the potential energy of a point charge q at r. V(r) is the electric potential at the point r.

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Electric potential The units of electric potential must be energy/charge. Our unit is the volt; 1 V = J/C = joules/coulombs. Conversely, eV (the electron volt) is a unit of energy because it is charge (e) × potential (V). F q (r) is the electric force on a point charge q located at r. E(r) is the electric field at the point r. U(r) is the potential energy of a point charge q at r. V(r) is the electric potential at the point r.

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Electric potential Example: An electron accelerates from rest through an electric potential of 2 V. What is the final speed of the electron?

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Electric potential Example: An electron accelerates from rest through an electric potential of 2 V. What is the final speed of the electron? Answer: The electron acquires 2 eV kinetic energy. We have and since the mass of the electron is m e = 9.1 × 10 –31 kg, the speed is

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Electric potential The electric potential V(r) at a point r, due to a system of n point charges q i located at r i, is The electric potential V(r) at a point r, due to a continuous charge density ρ(r), is

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Electric potential Example 1: What is the electric potential at a point (x,y,z) due to a dipole made of charges q at (0,0,d/2) and –q at (0,0, –d/2)? z –d/2 d/2 (x,y,z)(x,y,z)

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Electric potential Example 1: What is the electric potential at a point (x,y,z) due to a dipole made of charges q at (0,0,d/2) and –q at (0,0, –d/2)? Answer: z –d/2 d/2 (x,y,z)(x,y,z) r+r+ r–r–

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Electric potential Example 2: A rod with a uniform charge density λ lies along the z-axis between (0,0,–L/2) and (0,0,L/2). Let’s calculate the electric potential V(y) at a point (0,y,0) on the y-axis. z y –L/2 L/2

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Electric potential Example 2: A rod with a uniform charge density λ lies along the z-axis between (0,0,–L/2) and (0,0,L/2). Let’s calculate the electric potential V(y) at a point (0,y,0) on the y-axis. z y –L/2 L/2

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Halliday, Resnick and Krane, 5 th Edition, Chap. 28, Prob. 4: The electric field inside a nonconducting, uniform sphere of radius R has magnitude E = qr/4πε 0 R 3, where q is the total charge on the sphere and r is the distance from its center. (a) Given V(0) = 0, compute V(r). (b) What is the difference in electric potential between the surface of the sphere and its center? (c) Given V(∞) = 0, show that V(r) = q(3R 2 –r 2 )/8πε 0 R 3 inside the sphere. Why does this answer differ from the answer in part (a)?

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Halliday, Resnick and Krane, 5 th Edition, Chap. 28, Prob. 4: The electric field inside a nonconducting, uniform sphere of radius R has magnitude E = qr/4πε 0 R 3, where q is the total charge on the sphere and r is the distance from its center. (a) Given V(0) = 0, compute V(r). Answer: Symmetry dictates that E is radial. So we have to integrate E(r) = qr/4πε 0 R 3 with respect to dr and, if necessary, add a constant to make V(0) = 0. Thus and C = 0 yields V(0) = 0.

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Halliday, Resnick and Krane, 5 th Edition, Chap. 28, Prob. 4: The electric field inside a nonconducting, uniform sphere of radius R has magnitude E = qr/4πε 0 R 3, where q is the total charge on the sphere and r is the distance from its center. (b) What is the difference in electric potential between the surface of the sphere and its center? Answer: The question asks what is V(R) – V(0), and the answer is – q/8πε 0 R.

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Halliday, Resnick and Krane, 5 th Edition, Chap. 28, Prob. 4: The electric field inside a nonconducting, uniform sphere of radius R has magnitude E = qr/4πε 0 R 3, where q is the total charge on the sphere and r is the distance from its center. (c) Given V(∞) = 0, show that V(r) = q(3R 2 –r 2 )/8πε 0 R 3 inside the sphere. Why does this answer differ from the answer in part (a)? Answer: Outside the sphere, the potential must be V(r) = q/4πε 0 r, because the sphere looks just like a point charge q, and then we also satisfy V(∞) = 0. This outside potential must match the inside potential V(r) = q(3R 2 –r 2 )/8πε 0 R 3 at r = R and, indeed, both expressions yield V(R) = q/4πε 0 R. So V(r) is the same potential as in (a); they differ only by a constant so that in (a) we have V(0) = 0 and here we have V(0) = 3q/8πε 0 R.